CBSE 2024 · Region 2 · Set 1 · Q35 · 5 marks
If $\displaystyle \mathrm{A}_{1}$ denotes the area of region bounded by $\displaystyle \mathrm{y}^{2}=4 \mathrm{x}, \mathrm{x}=1$ and x -axis in the first quadrant and $\displaystyle \mathrm{A}_{2}$ denotes the area of region bounded by $\displaystyle \mathrm{y}^{2}=4 \mathrm{x}, \mathrm{x}=4$, find $\displaystyle \mathrm{A}_{1}: \mathrm{A}_{2}$.
Marking-scheme solution
$$\mathrm{A}_{1}=\text { Area }(\text { region } \mathrm{OABO})=\int_{0}^{1} 2 \sqrt{\mathrm{x}} d \mathrm{x}=\left.2\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\dfrac{3}{2}}\right]\right|_{0} ^{1}=\frac{4}{3}$\displaystyle \mathrm{A}_{2}=$ Area $\displaystyle ($ region ODEO $\displaystyle )=2 \int_{0}^{4} 2 \sqrt{\mathrm{x}} d \mathrm{x}=4 \times \frac{2}{3}\left[2^{3}\right]=\frac{64}{3}$
$\displaystyle \mathrm{A}_{1}: \mathrm{A}_{2}=\frac{4}{3}: \frac{64}{3}=1: 16$
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